The Tale of Point P

$x^2 + y^2 = 194$
$AB^2 = 1+x^2$
$\triangle PHB \sim \triangle BOA$
$\dfrac{PB}{BA} = \dfrac{BH}{AO}$

$PB = y , BH = \dfrac12 AB$

$\dfrac12 AB^2 =y$
$1 + x^2 =2y$
$195-y^2 =2y$
$y^2 + 2y - 195 = 0$
$y = 13 \Rightarrow x=5$
$\text{area} [AOBP] = \dfrac12 x(1+y) = \boxed{35}$ .

The two triangles form a trapezoid that has 2 right angles at point $O$ and point $B$ , and $AP = PB$ . Therefore, point $A$ can act as a focus of a parabola with an x-axis as its directrix, as shown below.

In a parabolic graph, for the focus point $F$ and point $P$ on the graph, the length of $FP$ will always equal the distance from point $P$ to the directrix, a line that has a distance to the parabola's vertex equal to the distance from the vertex to the focus point. In other words, the vertex of the parabola will be the midpoint on the distance between the focus and the directrix.

As a result, the vertex of the parabola is point $(0, \dfrac{1}{2})$ which is the midpoint between point $A$ and point $O$ , and so the focus distance $f$ is $\dfrac{1}{2}$ .

Now we can formulate the parabola's equation as:

$4fy = x^2 + C$

Plugging in $f = \dfrac{1}{2}$ and the y-intercept $(0, \dfrac{1}{2})$ , we will get:

$2y = x^2 + 1$

Now the length of $OP$ = $\sqrt{194}$ = $\sqrt{x^2 + y^2}$ .

$194 = (2y - 1) + y^2$

$y^2 + 2y - 195 = 0$

$(y - 13)(y + 15) = 0$

Thus, $y = 13$ , and $x^2 = 25 ; x = 5$ .

That is, $PB = 13$ and $OB = 5$ .

Then the area of the trapezoid $AOBP$ = $\dfrac{1}{2}\times(1+13)\times 5$ = $\boxed{35}$ .

$Let P(Xp,Yp).~~~~~~~~~~\\ P ~ is ~ on ~ \odot~~ Xp^2+Yp^2=194.......(I)\\ Also~AP^2=PB^2 ~~~~~ \\ \implies Xp^2+(Yp - 1)^2=Yp^2~~~~~\\ \implies~Xp^2= - 1+2Yp.......(II)\\ \therefore~putting ~ (II) ~ in~(I). ~and ~ solving,~\\ we ~get ~ Yp^2+2Yp - 195=0.\\ \text{But on solving the quadratic, }\\ positive ~~ Yp=13, \\ and ~ from~ (II) ~ Xp=5. ~~ Xp\neq - 5.\\ Required ~area=\frac 1 2 *(1+13)*5=35.$

If OP = sqrt (194) then OB^2 + BP^2 = 194.

The only possible integer values are 5 and 13.

P(13,5) doesn't work because BP would be too short compared to AP.

Therefore P is the point (5,13). AP = sqrt(5^2 + 12^2)= sqrt(169) = 13 = BP.

Let D be the point at (0,13). Then rectangle OBPD has an area of 5 X 13 = 65.

Triangle APD has an area =0 .5 12 5 = 30.

The difference is the required quadrilateral, 65 - 30 = 35.